Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#define x first
#define y second

typedef long long ll;
typedef std::pair<int,int> pii;

const int N=250010,MOD=1000000007,g0=500002,g1=500003;

pii operator * (pii A,pii B){return pii((ll)A.x*B.x%MOD,(ll)A.y*B.y%MOD);}
pii operator + (pii A,pii B){return pii(((A.x+B.x)%MOD+MOD)%MOD,((A.y+B.y)%MOD+MOD)%MOD);}
pii operator - (pii B){return pii(-B.x,-B.y);}

ll inv(int x){return (x==1 || x==-1)?x:(-(MOD/x)*inv(MOD%x)%MOD);}

pii t0[N*2],*pow;

void initialize()
{
	pow=t0+N;

	pii tmp=pii(g0,g1);

	pow[0]=pii(1,1);
	for(int i=1;i<N;i++)
		pow[i]=pow[i-1]*tmp;

	tmp=pii((inv(g0)+MOD)%MOD,(inv(g1)+MOD)%MOD);
	for(int i=0;i<N;i++)
		pow[-i-1]=pow[-i]*tmp;
}

char s[N];
int ind[N];
pii val[N],sum[N];
int n;

ll ans;

void prework()
{
	int p=0;
	for(int i=1;i<=n;i++)
	{
		ind[i]=p,val[i]=pii(0,0);

		if(s[i]=='<')p--;
		else if(s[i]=='>')p++;
		else if(s[i]=='+')val[i]=pow[p];
		else val[i]=-pow[p];
	}

	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+val[i];
}

std::map<pii,int> G;

void solve()
{
	ans=0;

	pii x,y;

	for(int i=n;i;i--)
	{
		G[sum[i]]++;

		x=sum[n]*pow[ind[i]];
		y=sum[i-1]+x;

		ans+=G[y];
	}
}

int main()
{
	initialize();

	scanf("%d",&n);
	scanf("%s",s+1);

	prework();
	solve();

	printf("%lld\n",ans);

	return 0;
}